Hi. I'd like to port a Perl function that does something I don't
know how to do in Python. (In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? If not, how would
one code a similar functionality in Python? Basically the problem
is to split one list into two according to some criterion.
TIA!
Kynn
--
NOTE: In my address everything before the first period is backwards;
and the last period, and everything after it, should be discarded.
10  1166 
Hi. I'd like to port a Perl function that does something I don't
know how to do in Python. (In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? If not, how would
one code a similar functionality in Python? Basically the problem
is to split one list into two according to some criterion.
This function will take a list of integers and modify it in place such
that it removes even integers. The removed integers are returned as a
new list (disclaimer: I'm 100% sure it can be done better, more
optimized, etc, etc):
def mod( alist ):
old = alist[:]
ret = [ ]
for i in old:
if i % 2 == 0:
ret.append( alist.pop( alist.index( i ) ) )
return ret
x = range(10,20)
print x
r = mod( x )
print r
print x
HTH,
Daniel
--
Psss, psss, put it down! - http://www.cafepress.com/putitdown
>This function will take a list of integers and modify it in place such
>that it removes even integers. The removed integers are returned as a
new list
<snip>
Great!
Thanks!
kynn
--
NOTE: In my address everything before the first period is backwards;
and the last period, and everything after it, should be discarded.
On Jun 7, 2:42*pm, "Daniel Fetchinson" <fetchin...@goo glemail.com>
wrote:
Hi. *I'd like to port a Perl function that does something I don't
know how to do in Python. *(In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. *Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? *If not, how would
one code a similar functionality in Python? *Basically the problem
is to split one list into two according to some criterion.
This function will take a list of integers and modify it in place such
that it removes even integers. The removed integers are returned as a
new list (disclaimer: I'm 100% sure it can be done better, more
optimized, etc, etc):
def mod( alist ):
* * old = alist[:]
* * ret = [ ]
* * for i in old:
* * * * if i % 2 == 0:
* * * * * * ret.append( alist.pop( alist.index( i ) ) )
* * return ret
x = range(10,20)
print x
r = mod( x )
print r
print x
HTH,
Daniel
--
Psss, psss, put it down! -http://www.cafepress.c om/putitdown
def mod( alist ):
return [ alist.pop( alist.index( x ) ) for x in alist if x % 2 ==
0 ]
alist = range(10,20)
blist = mod( alist )
print alist
print blist
The same thing with list comprehensions.
kj wrote:
Hi. I'd like to port a Perl function that does something I don't
know how to do in Python. (In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
The two solutions thus far use the .index() method, which itself runs in
O(n) time. This means that the provided functions run in O(n^2) time,
which can be a problem if the list is big. I'd go with this:
def partition(alist , criteria):
list1, list2 = [], []
for item in alist:
if criteria(item):
list1.append(it em)
else:
list2.append(it em)
return (list1, list2)
def mod(alist, criteria=lambda x: x % 2 == 0):
alist[:], blist = partition(alist , criteria)
return blist
>>partition(ran ge(10), lambda x: x % 2 == 0)
([0, 2, 4, 6, 8], [1, 3, 5, 7, 9])
>>l=range(10)
mod(l)
[1, 3, 5, 7, 9]
>>l
[0, 2, 4, 6, 8]
On Jun 7, 1:24*pm, kj <so...@987jk.co m.invalidwrote:
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. *Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? *If not, how would
one code a similar functionality in Python? *Basically the problem
is to split one list into two according to some criterion.
If you want to avoid side-effects completely, return two lists, the
list of matches and the list of non-matches. In this example,
partition creates two lists, and stores all matches in the first list,
and mismatches in the second. (partition assigns to the associated
element of retlists based on False evaluating to 0 and True evaluating
to 1.)
def partition(lst, ifcond):
retlists = ([],[])
for i in lst:
retlists[ifcond(i)].append(i)
return retlists[True],retlists[False]
hasLeadingVowel = lambda x: x[0].upper() in "AEIOU"
matched,unmatch ed = partition("The quick brown fox jumps over the lazy
indolent dog".split(),
hasLeadingVowel )
print matched
print unmatched
prints:
['over', 'indolent']
['The', 'quick', 'brown', 'fox', 'jumps', 'the', 'lazy', 'dog']
-- Paul
On Jun 8, 6:05 am, eat...@gmail.co m wrote:
On Jun 7, 2:42 pm, "Daniel Fetchinson" <fetchin...@goo glemail.com>
wrote:
Hi. I'd like to port a Perl function that does something I don't
know how to do in Python. (In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? If not, how would
one code a similar functionality in Python? Basically the problem
is to split one list into two according to some criterion.
This function will take a list of integers and modify it in place such
that it removes even integers. The removed integers are returned as a
new list (disclaimer: I'm 100% sure it can be done better, more
optimized, etc, etc):
def mod( alist ):
old = alist[:]
ret = [ ]
for i in old:
if i % 2 == 0:
ret.append( alist.pop( alist.index( i ) ) )
return ret
x = range(10,20)
print x
r = mod( x )
print r
print x
HTH,
Daniel
--
Psss, psss, put it down! -http://www.cafepress.c om/putitdown
def mod( alist ):
return [ alist.pop( alist.index( x ) ) for x in alist if x % 2 ==
0 ]
alist = range(10,20)
blist = mod( alist )
print alist
print blist
The same thing with list comprehensions.
Not the same. The original responder was careful not to iterate over
the list which he was mutating.
>>def mod(alist):
.... return [alist.pop(alist .index(x)) for x in alist if x % 2 == 0]
....
>>a = range(10)
print mod(a), a
[0, 2, 4, 6, 8] [1, 3, 5, 7, 9]
>>a = [2,2,2,2,2,2,2,2]
print mod(a), a
[2, 2, 2, 2] [2, 2, 2, 2]
# should be [2, 2, 2, 2, 2, 2, 2, 2] []
On Jun 7, 5:56*pm, John Machin <sjmac...@lexic on.netwrote:
On Jun 8, 6:05 am, eat...@gmail.co m wrote:
On Jun 7, 2:42 pm, "Daniel Fetchinson" <fetchin...@goo glemail.com>
wrote:
Hi. *I'd like to port a Perl function that does something I don't
know how to do in Python. *(In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. *Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? *If not, how would
one code a similar functionality in Python? *Basically the problem
is to split one list into two according to some criterion.
This function will take a list of integers and modify it in place such
that it removes even integers. The removed integers are returned as a
new list (disclaimer: I'm 100% sure it can be done better, more
optimized, etc, etc):
def mod( alist ):
* * old = alist[:]
* * ret = [ ]
* * for i in old:
* * * * if i % 2 == 0:
* * * * * * ret.append( alist.pop( alist.index( i ) ) )
* * return ret
x = range(10,20)
print x
r = mod( x )
print r
print x
HTH,
Daniel
--
Psss, psss, put it down! -http://www.cafepress.c om/putitdown
def mod( alist ):
* * return [ alist.pop( alist.index( x ) ) for x in alist if x % 2 ==
0 ]
alist = range(10,20)
blist = mod( alist )
print alist
print blist
The same thing with list comprehensions.
Not the same. The original responder was careful not to iterate over
the list which he was mutating.
>def mod(alist):
... * *return [alist.pop(alist .index(x)) for x in alist if x % 2 == 0]
...>>a = range(10)
>print mod(a), a
[0, 2, 4, 6, 8] [1, 3, 5, 7, 9]>>a = [2,2,2,2,2,2,2,2]
>print mod(a), a
[2, 2, 2, 2] [2, 2, 2, 2]
# should be [2, 2, 2, 2, 2, 2, 2, 2] []
Alas, it appears my understanding of list comprehensions is
significantly less comprehensive than I thought =)
On Jun 8, 8:17 am, eat...@gmail.co m wrote:
On Jun 7, 5:56 pm, John Machin <sjmac...@lexic on.netwrote:
On Jun 8, 6:05 am, eat...@gmail.co m wrote:
On Jun 7, 2:42 pm, "Daniel Fetchinson" <fetchin...@goo glemail.com>
wrote:
Hi. I'd like to port a Perl function that does something I don't
know how to do in Python. (In fact, it may even be something that
is distinctly un-Pythonic!)
The original Perl function takes a reference to an array, removes
from this array all the elements that satisfy a particular criterion,
and returns the list consisting of the removed elements. Hence
this function returns a value *and* has a major side effect, namely
the target array of the original argument will be modified (this
is the part I suspect may be un-Pythonic).
Can a Python function achieve the same effect? If not, how would
one code a similar functionality in Python? Basically the problem
is to split one list into two according to some criterion.
This function will take a list of integers and modify it in place such
that it removes even integers. The removed integers are returned as a
new list (disclaimer: I'm 100% sure it can be done better, more
optimized, etc, etc):
def mod( alist ):
old = alist[:]
ret = [ ]
for i in old:
if i % 2 == 0:
ret.append( alist.pop( alist.index( i ) ) )
return ret
x = range(10,20)
print x
r = mod( x )
print r
print x
HTH,
Daniel
--
Psss, psss, put it down! -http://www.cafepress.c om/putitdown
def mod( alist ):
return [ alist.pop( alist.index( x ) ) for x in alist if x % 2 ==
0 ]
alist = range(10,20)
blist = mod( alist )
print alist
print blist
The same thing with list comprehensions.
Not the same. The original responder was careful not to iterate over
the list which he was mutating.
>>def mod(alist):
... return [alist.pop(alist .index(x)) for x in alist if x % 2 == 0]
...>>a = range(10)
>>print mod(a), a
[0, 2, 4, 6, 8] [1, 3, 5, 7, 9]>>a = [2,2,2,2,2,2,2,2]
>>print mod(a), a
[2, 2, 2, 2] [2, 2, 2, 2]
# should be [2, 2, 2, 2, 2, 2, 2, 2] []
Alas, it appears my understanding of list comprehensions is
significantly less comprehensive than I thought =)
It's nothing to do with list comprehensions, which are syntactical
sugar for traditional loops. You could rewrite your list comprehension
in the traditional manner:
def mod(alist):
ret = []
for x in alist:
if x % 2 == 0:
ret.append(alis t.pop(alist.ind ex(x))
return ret
and it would still fail for the same reason: mutating the list over
which you are iterating.
At the expense of even more time and memory you can do an easy fix:
change 'for x in alist' to 'for x in alist[:]' so that you are
iterating over a copy.
Alternatively, go back to basics:
>>def modr(alist):
.... ret = []
.... for i in xrange(len(alis t) - 1, -1, -1):
.... if alist[i] % 2 == 0:
.... ret.append(alis t[i])
.... del alist[i]
.... return ret
....
>>a = [2,2,2,2,2,2,2,2 ,2]
print modr(a), a
[2, 2, 2, 2, 2, 2, 2, 2, 2] []
>>>
HTH,
John
In <72************ *************** *******@f24g200 0prh.googlegrou ps.comJohn Machin <sj******@lexic on.netwrites:
>It's nothing to do with list comprehensions, which are syntactical
sugar for traditional loops. You could rewrite your list comprehension
in the traditional manner...
and it would still fail for the same reason: mutating the list over
which you are iterating.
I normally deal with this problem by iterating backwards over the
indices. Here's how I coded the function (in "Python-greenhorn
style"):
def cull(list):
culled = []
for i in range(len(list) - 1, -1, -1):
if not_wanted(list[i]):
culled.append(l ist.pop(i))
return culled
....where not_wanted() is defined elsewhere. (For my purposes at the
moment, the greater generality provided by making not_wanted() a
parameter to cull() was not necessary.)
The specification of the indices at the beginning of the for-loop
looks pretty ignorant, but aside from that I'm happy with it.
Kynn
--
NOTE: In my address everything before the first period is backwards;
and the last period, and everything after it, should be discarded. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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